6a+7b+8c+9d=100,
7b=100-6a-8c-9d,
7b=98+2+a-7a-c-7c-7d-2d,
b=14+2/7+a/7-a-c/7-c-d-2d/7,
b=(14-a-c-d)+(2+a-c-2d)/7.
We need to find a+2-(c+2d) such that the sum is divisible by 7.
For example, if a=9 then a+2=11; if c+2d=4 then the whole sum adds up to 7.
c=4-2d, so if d=1, c=2. Therefore one solution is (a,c,d)=(9,2,1).
b=14-(9+2+1)+1=3. So one solution is (a,b,c,d)=(9,3,2,1). (6a,7c,8c,9d)=(54,21,16,9)
CHECK: 54+21+16+9=100✔️
There are many other integer solutions, including negative values for the variables.
There are also many more solutions if fractions are permitted.